How to solve this absolute value inequality?

|x + 1| > x + 2

The way I think about it, if this was an equation, removing the absolute value puts a ± on the other side.

Since this is an inequality, the - side of that has to have the sign flipped.

Once you do that, then the solution, if any, would be the overlap of the two solutions.

So doing that here gets us to:

x + 1 > x + 2 and x + 1 < -(x + 2)

Solving for both, we get:

1 > 2 and x + 1 < -x - 2

FALSE and 2x < -3

FALSE and x < -3/2

Since one of the solutions was no solution, we can throw it out leaving only:

x < -3/2

as your solution. As a test, if I pick two numbers on either side, I should be able to predict the solutions.

I'll say -2 and 0 to be TRUE and FALSE:

|x + 1| > x + 2

|-2 + 1| > -2 + 2 and |0 + 1| > 0 + 2

|-1| > 0 and |1| > 2

1 > 0 and 1 > 2

TRUE and FALSE

So I was correct, so :

x < -3/2 is the solution

I ususally find a graph of an absolute value equation to be revealing. Subtract the right side to get

.. |x+1| - (x+2) > 0

The graph at the source link shows this to be true for x < -3/2.

If |x + 1| > x + 2, then either x + 1 > x + 2 (ALWAYS a false statement) or x + 1 < -(x + 2). Now:

x + 1 < -x - 2

2x + 1 < -2

2x < -3

x < -3/2. That's the final answer.

|x + 1| > x + 2

When x >= 1,

x + 1 > x + 2

x - x + 1 > x - x + 2

1 > 2 which is false for all x

When x < 1,

-(x + 1) > x + 2

-x - 1 > x + 2

-3 > 2x

x < -3/2

Final answer: x < -3/2